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Projectile Questions

  1. Sep 8, 2013 #1
    1. The problem statement, all variables and given/known data

    We did an experiment where we threw a tennis ball far and high.
    For the far toss:
    initial height=1.3m
    time it took to land=1.92sec
    distance traveled=23.4m

    For the high toss:
    initial height=1.37m
    time it took to land=2.61sec
    distance traveled=9.5m

    Now I have to calculate the maximum height for both tosses.
    Then I have to solve for the xmax at maximum height for the long toss.

    2. Relevant equations
    I know that I have to use delta y=VyoT-.5gt^2 for maximum height.


    3. The attempt at a solution
    I know the formula to solve for maximum height but there isn't any angle that can be used so I don't know what to do. I tried setting Vyo equals to 0 but that gave me a negative number. For the xmax, I know it is supposed to be less than half of the distance traveled because of the initial height but I don't know how to find it.
     
  2. jcsd
  3. Sep 8, 2013 #2
    If you only care about the max height, ignore the horizontal distance part. Focus solely on the vertical motion. The equation you have is good. What is known and unknown in it?
     
  4. Sep 8, 2013 #3

    TSny

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    Homework Helper
    Gold Member

    Hello, KingPapaya. Welcome to PF!

    Consider the far toss. You know the equation that relates y and t. Apply that equation for t = 1.92 s. What can you deduce from that equation?

    Repeat for the equation that relates x and t.

    [Oops, I see voko already replied while I was constructing my reply.]
     
  5. Sep 8, 2013 #4
    Delta y is y-yo so I have yo, which is 1.3m for far/long and 1.37m for high.
    I have t, which is 1 .92 sec for far/long and 2.61 sec for high.
    G is 9.8m/s^2.
    I'm supposed to be solving for y but I'm not sure what Vyo is
     
  6. Sep 8, 2013 #5
    Since the ball hits the ground, do you think you don't have ## \Delta y ##?
     
  7. Sep 8, 2013 #6
    I thought I have Δy because I have yo
     
  8. Sep 8, 2013 #7
    Thus you have everything in that equation - except ## v_{0y} ##. Can you find ## v_{0y} ## then?
     
  9. Sep 8, 2013 #8
    So Vyo=(Δy+.5gt^2)/t?

    Vyo=(y-1.3m+.5(9.8ms^2)(1.92sec))/1.92sec

    Wait so what do I do about y, since it is also unknown
     
  10. Sep 8, 2013 #9
    Where is the ball at t = 1.92 s?
     
  11. Sep 8, 2013 #10
    Oh so since it is on the ground, Δy is -1.3m because 0-1.3m?
     
  12. Sep 8, 2013 #11
    Correct!
     
  13. Sep 8, 2013 #12
    After plugging in the numbers to the equation the number I get for long and high are the maximum heights?
    Vyo=maximum height?
     
  14. Sep 8, 2013 #13
    ## v_{0y} ## is the initial vertical velocity.

    What is the vertical velocity at the max height?
     
  15. Sep 8, 2013 #14
    Would it be 0?
     
  16. Sep 8, 2013 #15
    That is correct as well. Now that you know the initial and the final velocities, and you know the acceleration as well, can you find the time?
     
  17. Sep 8, 2013 #16
    0=Vyo-gt?
    So T is (0-Vyo)/-g?
     
  18. Sep 8, 2013 #17
    Yes.
     
  19. Sep 8, 2013 #18
    Now what should I do with the time and Vyo solved for to get the maximum height?
    Will I have to use the original equation?
     
  20. Sep 8, 2013 #19
    Is there any reason you cannot?
     
  21. Sep 8, 2013 #20
    No.
    So after I do the same thing for the high toss, should I make a triangle?
     
    Last edited: Sep 8, 2013
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